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3 years ago

What is a proton?

Physics

2 answers:

eimsori [14]3 years ago

7 0

The answer is A. A positive particle inside the nucleus

kirill115 [55]3 years ago

3 0

The answer is A.
I hope this helps

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An elevator motor in a high-rise building can do 3500 kJ of work in 5 min. Find the power developed by the motor. Explain if you

Ilya [14]

Answer:

P = 11666.6 W

Explanation:

Given that,

Work done by the motor, W = 3500 kJ

Time, t = 5 min = 300 s

We need to find the power developed by the motor. Power developed is given by :

$P=\dfrac{E}{t}\\\\P=\dfrac{3500\times 10^3}{300}\\\\P=11666.7\ W$

So, the required power is 11666.6 W.

4 0

3 years ago

From what expression was the word radar derived

tiny-mole [99]

RAdio Direction And Ranging

7 0

3 years ago

A 48.0-kg astronaut is in space, far from any objects that would exert a significant gravitational force on him. He would like t

marusya05 [52]

Answer:

The astronaut is moving at a speed of 0.36m/s

Explanation:

Speed here corresponds to velocity

The astronaut's mass = 48kg

velocity of astronaut = ?

mass of socket = 0.72kg

velocity of socket = 5m/s

mass of the spanner = 0.8kg

velocity of spanner = 8m/s

change in time = 0.05 -0 = 0.05sec

mass of the mallet = 1.2kg

velocity of mallet = 6m/s

change in time = 9.9 -0 = 9.9sec

To find the astronaut velocity, we would calculate the total momentum which is the astronaut.

∑momentum (M) = ∑astronaut momentum

∑M = ∑astronaut M

∑astronaut M = M of socket + M of spanner + M of mallet

momentum = mass × velocity

(mass × velocity)of astronaut = (0.72×5) + (0.8×8) + (1.2×6)

48 × velocity of astronaut= 3.6 + 6.4 + 7.2

48 × velocity of astronaut= 17.2

velocity of astronaut = 17.2/48

velocity of astronaut = 0.36m/s

The astronaut is moving at a speed of 0.36m/s

5 0

3 years ago

A book is dropped from a window. It takes 5 seconds to reach the ground. What is its velocity after 2 seconds? What’s the veloci

ira [324]

Answer:

Explanation:

Initial velocity is 0. In the equation v = v0+at where v0 is the initial velocity of 0, we only have to fill in -9.8 for a and 2 for t to get the velocity after 2 seconds -19.6 m/s; after 5 seconds, when it hits the ground, a = -9.8 and t = 5 to give a velocity of -49 m/s. Gravity pulls down everything at the same rate, it doesn't matter whether we drop a feather or an elephant from the window!

3 0

3 years ago

At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

Bad White [126]

Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

$\displaystyle \theta =68^o$

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$

$\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}$

Now we compute the magnitude

$\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}$

The direction of the total force is given by

$\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558$

$\boxed{\displaystyle \theta =68^o}$

6 0

3 years ago